Graph this system of equations and solve. $-8x-y = -3$ $6x+2y = -4$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
Convert the first equation, $-8x-y = -3$ , to slope-intercept form. $y = -8 x + 3$ The y-intercept for the first equation is $3$ , so the first line must pass through the point $(0, 3)$ The slope for the first equation is $-8$ . Remember that the slope tells you rise over run. So in this case for every $8$ positions you move down (because it's negative) $1$ position to the right. $8$ positions down from $(0, 3)$ is $(1, -5)$ Graph the blue line so it passes through $(0, 3)$ and $(1, -5)$ Convert the second equation, $6x+2y = -4$ , to slope-intercept form. $y = -3 x - 2$ The y-intercept for the second equation is $-2$ , so the second line must pass through the point $(0, -2)$ The slope for the second equation is $-3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) $1$ position to the right. $3$ positions down from $(0, -2)$ is $(1, -5)$ Graph the green line so it passes through $(0, -2)$ and $(1, -5)$ The solution is the point where the two lines intersect. The lines intersect at $(1, -5)$.